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A result by Sy Friedman in his book "fine structure and class forcing", is that, assume $0^\sharp$ exists, there exists a real number R such that the ordinals admissible in R (called $\Lambda(R)$) is exactly recursively inaccessible ordinals.

Sy Friedman proved that, if R belongs to a set generic extension of $L$, then $\Lambda(R)$ and $\Lambda(0)$ are only different at some initial segment of ordinals, so the result contradicts $V=L$. And it seems that this result is weaker than the existence of $0^\sharp$.

So it make me feel curious that, is this contradiction essential in some sense? As far as I know every large cardinal axiom weaker than $0^\sharp$ is consistent with $V=L$. If this result is stronger than any such axioms, then we will have a very interesting new intermediate level of consistency strength.

So, what is the exact strength of this result?

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